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2013 AMC 10A Problems/Problem 24

Revision as of 13:18, 16 December 2020 by Mathpro12345 (talk | contribs) (→‎Solution 2)

Problem

Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?

$\textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900$

Solution 1

Let us label the players of the first team $A$, $B$, and $C$, and those of the second team, $X$, $Y$, and $Z$.

$\textbf{1}$. One way of scheduling all six distinct rounds could be: 

Round 1---->$AX$ $BY$ $CZ$
Round 2---->$AX$ $BZ$ $CY$
Round 3---->$AY$ $BX$ $CZ$
Round 4---->$AY$ $BZ$ $CX$
Round 5---->$AZ$ $BX$ $CY$
Round 6---->$AZ$ $BY$ $CX$

The above mentioned schedule ensures that each player of one team plays twice with each player from another team. Now you can generate a completely new schedule by permutating those $6$ rounds and that can be done in $6!=720$ ways.

$\textbf{2}$. One can also make the schedule in such a way that two rounds are repeated.

(a) 

Round 1---->$AX$ $BZ$ $CY$
Round 2---->$AX$ $BZ$ $CY$
Round 3---->$AY$ $BX$ $CZ$
Round 4---->$AY$ $BX$ $CZ$
Round 5---->$AZ$ $BY$ $CX$
Round 6---->$AZ$ $BY$ $CX$

(b) 

Round 1---->$AX$ $BY$ $CZ$
Round 2---->$AX$ $BY$ $CZ$
Round 3---->$AY$ $BZ$ $CX$
Round 4---->$AY$ $BZ$ $CX$
Round 5---->$AZ$ $BX$ $CY$
Round 6---->$AZ$ $BX$ $CY$

As mentioned earlier any permutation of (a) and (b) will also give us a new schedule. For both (a) and (b) the number of permutations are $\frac{6!}{2!2!2!}$ = $90$


So the total number of schedules is $720+90+90$ =$\boxed{\textbf{(E)} 900}$.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc10a/358

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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